User talk:Hyp cos/Dimensional Array Notation
Extension of DAN: {0({0}),1} = {0(n),1}, {0(X),1} = {0,,...,,1} (X ,'s) Limit: {0({0(...),1}),1}. AarexTiao 21:53, November 18, 2013 (UTC) Message Hyp cos, please expand this notation? AarexTiao 12:48, November 22, 2013 (UTC) Oh! I've just realized that I created some notations very complex but doesn't get much further. I think this just get \(\psi(\psi_{\chi(M^{\omega^\omega})}(0))\) or a little above. So I want to remake them. hyp$hyp?cos&cos (talk) 02:33, November 23, 2013 (UTC) :Well? {0{1*}1} has level \(\psi(\psi_{\Xi(\omega,0)}(0))\) :{0{n*}1} has level \(\psi(\psi_{\Xi(\omega^n,0)}(0))\) :{0 :Definitions? LittlePeng9 (talk) 08:28, July 5, 2014 (UTC) Analysis I can believe in your comparisons here, they look pretty reasonable. But what if we compare it to CoC expressions and Taranovsky's notation? Ikosarakt1 (talk ^ ) 19:10, July 16, 2014 (UTC) My intuition is: not even close to either ofmentioned notations. LittlePeng9 (talk) 19:38, July 16, 2014 (UTC) I don't understand how they work in the recursive ordinal way. And, which one is stronger? hyp$hyp?cos&cos (talk) 14:17, July 17, 2014 (UTC) :I definitely believe that the proof theoretic ordinal for CoC is larger than anything Taranovsky reaches, although Taranovsky suggests that it may be otherwise in his latest analysis. For understanding Taranovsky's notation, you should work through the systems defined in his "Degrees of Recursive Inaccessibility" and "Degrees of Reflection" first, as these are easier to understand. The "Degrees of Reflection" version may already be stronger than your notation. Deedlit11 (talk) 08:42, July 18, 2014 (UTC) ::My guess is that the notation in "Degrees of Reflection" is as strong as PNAN. -- ☁ I want more ⛅ 13:06, July 18, 2014 (UTC) I guess \(C(0,\beta)=\beta+1\), and \(C(\alpha+1,\beta)\) is the limit of sequence \(C(\alpha,\beta)\), \(C(\alpha,C(\alpha,\beta))\), \(C(\alpha,C(\alpha,C(\alpha,\beta)))\), etc. am I right? hyp$hyp?cos&cos (talk) 15:10, July 18, 2014 (UTC) :It looks like that, but it works only up to \varepsilon_0 . Ikosarakt1 (talk ^ ) 15:34, July 18, 2014 (UTC) :But why \(\varepsilon_0\) has degree \(\Omega\) and \(C(\Omega,0)=\varepsilon_0\)? (Here the \(C(\Omega,0)\) is standard form) hyp$hyp?cos&cos (talk) 15:55, July 18, 2014 (UTC) :Because in C(a,b), "a" represents degree I think. Ikosarakt1 (talk ^ ) 18:56, July 18, 2014 (UTC) ::\(\varepsilon_0\) is \(C(0, C(\Omega, 0))\). \(C(\Omega, 0)\) is basically \(\Omega\) in our usual ordinal collapsing notations. ::In the notation for C(a, b, c) in "Degrees of Recursive Inaccesibility", a is the admissibility degree, b is the inductive level, and c is an ordinal that we take C(a, b, c) to be larger than. For the "Degrees of Reflection" notation, \(C(\Omega a + b, c)\) is C(a, b, c) from the previous notation, so \(C(\Omega, 0) = C(1, 0, 0)\), or the smallest ordinal of admissibility degree 1. Deedlit11 (talk) 21:21, July 18, 2014 (UTC) ::No, what I said is the n=1 system in "Ordinal Notation System for Second Order Arithmetic". Then why \(C(\Omega_1,0)=\varepsilon_0\)? hyp$hyp?cos&cos (talk) 01:07, July 19, 2014 (UTC) :::Ah, okay. For \(a < \Omega_1\), \(C(a, b) = b + \omega^a\). Remember the comparison of expressions is to write each expression in postfix form and compare lexicographically with \(C < 0 < \Omega_1\) (Keep in mind that the order of the variables is reversed, which is confusing.). Note that any expression composed of 0's and C's will have to start with a 0, so any such expression must be less than \(0 \Omega_1 C\). So \(C(\Omega_1, 0)\) must be greater than anything generated using C starting from 0, so it must be at least \(\varepsilon_0\). To show that it is exactly \(\varepsilon_0\), we must show that any standard expression with at least one \(\Omega_1\) must be greater than \(0 \Omega_1 C\). To be less than \(0 \Omega_1 C\), an expression must start with \(0 0\). So the task is to show that there is no standard expression starting with \(0 0\) and containing at least one \(\Omega_1\). I admit I don't see how to do that right now. Deedlit11 (talk) 04:00, July 19, 2014 (UTC) ::So is it "expression comparisons" (not "degrees") that decide the result of an expression? hyp$hyp?cos&cos (talk) 08:42, July 19, 2014 (UTC) ::: Yes, that is the way he defines it on the page. However, he does mention certain expressions equaling certain nonrecursive ordinals, so perhaps he has an alternative description of his notation that uses collapsing. Also, for n = 1, the notation is equivalent to the "Bachmann-Howard ordinal" notation further up the page, so for that at least we can use the "degree" definition. Deedlit11 (talk) 04:29, July 22, 2014 (UTC) Dimensional Array Notation using 5 symbols! If R function have 5 symbols then \(\{n*\}\) has changed to \(,\{n\}\)! There is 6 symbols: R, {, }, command, number, and asterisk. Now there is 5 symbols: R, {, }, command, and number. AarexTiaokhiao 22:12, August 6, 2014 (UTC) Stage cardinals did not reach D Me and alemagno12 analysised. *\(I_n\) = \(\Pi^0_n\) *\(M\) = \(\Pi^1_0\) *\(K\) = \(\Pi^1_1\) *\(U\) (Ultimate cardinal) = \(\Pi^1_2\) *Stage-5 = \(\Pi^1_3\) *\(S\) (Stage function) = \(\Sigma^2_0\) *\(T\) (Stage cardinal) = \(\Pi^2_0\) *\(T2\) (Stage stage cardinal) = \(\Pi^2_1\) *\(T3\) (Stage stage stage cardinal) = \(\Pi^2_2\) *Superstage cardinal = \(\Sigma^3_0\) *Limit of stage cardinal systems = \(\Pi^3_0\) AarexWikia04 - 23:23, August 4, 2016 (UTC) : Well, after analysing again, it seems that \(\Pi^2_1\)is the limit of stage cardinals. ---- From the googol and beyond -- 19:21, August 5, 2016 (UTC) :: Show me AarexWikia04 - 19:33, August 5, 2016 (UTC) * Supersrtage-1 cardinal \(\Pi^2_1\) * Superstage-2 cardinal \(\Pi^2_2\) * Superstage cardinal \(\Pi^3_2\) * Duperstage cardinal \(\Pi^3_0\) * Duperstage-1 cardinal \(\Pi^3_1\) * Truperstage-1 cardinal \(\Pi^4_1\) * Diagonalizer of n-erstage cardinal notation= D Bubby3 (talk) 22:22, November 26, 2016 (UTC)